Question

ACID-BASE TITRATIONS HOMEWORK Use the data in Option A to answer questions 1.5 1. Write a balanced equation for the lanced equation for the reaction of the acid and base indicated. CH₃CH2COOH(aq) + NOOH(aq) + CH₂CH₂COONa(aq) + H₂O(e) 2. Determine the volume of titrant added at the equivalence point. CH3CH₂ Cootrag) + NaOH(aa) CH₃CH aded at the equivalence point. SHOW ALL WORK! 19) + NaOH (aa) + CH₃ CH₂ COONacaq) + H2Ocel 3. Determine the pH of the solution at the equivalence point (HINT: Consider the products of the reaction. One of the products should be a salt having an ion which will undergo hydrolysis. Determine the concentration of this ion, and write an equation showing the reaction of the ion with water. Then, use an inital, change, final chart in order to work the problem. Refer to Module 14B, Learning Roals 7.8 and 9 of your CHEM 1020 Handbook for more help. Also, you may use your laminated data sheet for helpful information.) hydrox

Answer 1

1) First, we write the reaction

CH3CH2COOH + NaOH → CH3CHCOONa+H2O

2) Using the molarity equation we have:

n(NaOH) = [NaOH (Volumeof NaOH)

Where HA is the propionic acid.

At the equivalence point the moles of acid and the moles of base are the same.

(VHA) 0,100M) = (20mL) 0,100M)

$(V_{HA}) = \tfrac{(0,100M)(20mL)}{(0,100M)}$

The volume of titrant added at equivalence point should be 20mL.

3) At this point, the propanoate ion will undergo hydrolysis.

$CH_{3}CH_{2}COO^{-} + H_{2}O \rightleftharpoons CH_{3}CH_{2}COOH + OH^{-}$

Now, we find the propanoate concentration at the equivalence point volume, that is 40 mL.

n(A) = 2m Mol

$[A^{-}]=\frac{2.10^{-3}}{0,04L}$

$[A^{-}]=0,05M$

Where $A^{-}$ is the propanoate ion.

Now, we write the equilibrium expression.

$Kb=\frac{[OH^{-}][HA]}{[A^{-}]}$

the tabulated kb for propanoate is  $7,856.10^{-10}$.

 The concentration of  will be the same

$Kb=\frac{[OH^{-}]^{2}}{[A^{-}]}$

Now, we determine the [OH-] and then we can find the pH of the solution

$[OH^{-}]= \sqrt{Kb[A^{-}]}$

$[OH^{-}]= \sqrt{(7,586.10^{-10})(0,05)}$

M

Now, we applied -log to find the pOH

if...

4) the titration curve will start at high pH values ​​since initially there will be an excess of NaOH.

Then the value will begin to decrease until reaching the equivalence point (20ml).

And then it will reach a constant value in excess acid.

14Strong Base - Weak Acid pH Volume of acid added